Single Choice

The wavelength of the first spectral line in the Balmer series of hydrogen atom is $$6561 $$$$\mathring{A}$$. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

A$$1212$$ $$\mathring{A}$$
Correct Answer
B$$1640$$ $$\mathring{A}$$
C$$2430$$ $$\mathring{A}$$
D$$4687$$ $$\mathring{A}$$

Solution

For balmer series:
$$\displaystyle \frac{1}{\lambda}= R(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})$$

$$\displaystyle \frac{1}{6561}= R(\frac{1}{4}- \frac{1}{9}) = \frac{5R}{36}$$

$$\displaystyle \frac{1}{\lambda}= 4R( \frac{1}{4}- \frac{1}{16})$$

$$ \lambda = 1212\ \mathring{A}$$

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